Asked by Maria
Glass has a density of about 2500 Kg/m^3. fresh water has a density of 1000 kg/m^3. if a 5 kg block of glass is sitting on the bottom of a freshwater fish tank, what is the normal force on the floor of the tank?
30N is what my professor says is correct, but im confused how?
30N is what my professor says is correct, but im confused how?
Answers
Answered by
bobpursley
weight on bottom= mg-weightwaterdisp
= 5g-1g/cm^3*volume
but volume of the glass=5/2.5 =2cm^3
normal force=5*9.8-1g/cm^2 *2*9.8
normal forceabout 50-20=30N
= 5g-1g/cm^3*volume
but volume of the glass=5/2.5 =2cm^3
normal force=5*9.8-1g/cm^2 *2*9.8
normal forceabout 50-20=30N
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